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Marizza181 [45]
3 years ago
6

PLEASE I NEED HELP!

Mathematics
2 answers:
N76 [4]3 years ago
4 0

Answer:Complete the statements.

When r = 2, the first expression is . 6

When r = 2, the second expression is 6.

When r = 5, the first expression is 21.

When r = 5, the second expression is .

The expressions are not equivalent. 12

Step-by-step explanation:

i got it right on e d g e n u i t y

Step-by-step explanation:

NARA [144]3 years ago
3 0
Fourth one i think (the way u typed the q is confusing)
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Help please need those answers
icang [17]

We know: The sum of the measures of the angles of a triangle is equal 180°.

We have: m∠A =65°, m∠B =  (3x - 10)° and m∠C = (2x)°.

The equation:

65 + (3x - 10) + 2x = 180

(3x + 2x) + (65 - 10) = 180

5x + 55 = 180        <em>subtract 55 from both sides</em>

5x = 125          <em>divide both sides by 5</em>

x = 25

m∠B =  (3x - 10)° → m∠B =  (3 · 25 - 10)° = (75 - 10)° = 65°

m∠C = (2x)° → m∠C = (2 · 25)° = 50°

<h3>Answer: x = 25, m∠B = 65°, m∠C = 50°</h3>
8 0
3 years ago
Please help this is due today
prisoha [69]

Answer:

x=28°

y=152°

Step-by-step explanation:

x=28°

y=152°

3 0
2 years ago
Write the equation in slope-intercept form. 4x+3y=18
Helen [10]

Answer:

y = -4/3x + 6

Step-by-step explanation:

1. 3y - 4x + 3y = 18 - 3y

2. 4x = -3y + 18

3. 18 - 4x = -3y + 18 - 18

4. <u>-</u><u>3</u><u>/</u><u> </u><u>-3y = 4x - 18</u><u> </u><u>/</u><u>3</u>

5. y = -4/3x + 6

6 0
2 years ago
HELP CALCULUS 25 POINTS
shepuryov [24]

Answer:

A.) Max at x = 6 and Min at x = -6

Step-by-step explanation:

We say that f(x) has a relative (or local) maximum at x=c if f(x)≤f(c) f ( x ) ≤ f ( c ) for every x in some open interval around x=c . We say that f(x) has an absolute (or global) minimum at x=c if f(x)≥f(c) f ( x ) ≥ f ( c ) for every x in the domain we are working on.

4 0
3 years ago
A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min
sesenic [268]
A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})
\implies A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\dfrac t4\right)\dfrac{\text{lb}}{\text{gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}
A'(t)+\dfrac1{100}A(t)=6+3\sin\dfrac t4

We're given that A(0)=30. Multiply both sides by the integrating factor e^{t/100}, then

e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=6e^{t/100}+3e^{t/100}\sin\dfrac t4
\left(e^{t/100}A(t)\right)'=6e^{t/100}+3e^{t/100}\sin\dfrac t4
e^{t/100}A(t)=600e^{t/100}-\dfrac{150}{313}e^{t/100}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+C
A(t)=600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+Ce^{-t/100}

Given that A(0)=30, we have

30=600-\dfrac{150}{313}\cdot25+C\implies C=-\dfrac{174660}{313}\approx-558.02

so the amount of salt in the tank at time t is

A(t)\approx600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)-558.02e^{-t/100}
3 0
3 years ago
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